# 12-240/Classnotes for Tuesday November 15

${\displaystyle A={\begin{pmatrix}0&2&4&2&2\\4&4&4&8&0\\8&2&0&10&2\\6&3&2&9&1\end{pmatrix}}}$.
Solution. Using (invertible!) row/column operations we aim to bring ${\displaystyle A}$ to look as close as possible to an identity matrix:
 Do Get 1. Bring a ${\displaystyle 1}$ to the upper left corner by swapping the first two rows and multiplying the first row (after the swap) by ${\displaystyle 1/4}$. ${\displaystyle {\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\8&2&0&10&2\\6&3&2&9&1\end{pmatrix}}}$ 2. Add ${\displaystyle (-8)}$ times the first row to the third row, in order to cancel the ${\displaystyle 8}$ in position 3-1. ${\displaystyle {\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&-6&-8&-6&2\\6&3&2&9&1\end{pmatrix}}}$ 3. Likewise add ${\displaystyle (-6)}$ times the first row to the fourth row, in order to cancel the ${\displaystyle 6}$ in position 4-1. ${\displaystyle {\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}}}$ 4. With similar column operations (you need three of those) cancel all the entries in the first row (except, of course, the first, which is used in the canceling). ${\displaystyle {\begin{pmatrix}1&0&0&0&0\\0&2&4&2&2\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}}}$ 5. Turn the 2-2 entry to a ${\displaystyle 1}$ by multiplying the second row by ${\displaystyle 1/2}$. ${\displaystyle {\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}}}$ 6. Using two row operations "clean" the second column; that is, cancel all entries in it other than the "pivot" ${\displaystyle 1}$ at position 2-2. ${\displaystyle {\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}}}$ 7. Using three column operations clean the second row except the pivot. ${\displaystyle {\begin{pmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}}}$ 8. Clean up the row and the column of the ${\displaystyle 4}$ in position 3-3 by first multiplying the third row by ${\displaystyle 1/4}$ and then performing the appropriate row and column transformations. Notice that by pure luck, the ${\displaystyle 4}$ at position 4-5 of the matrix gets killed in action. ${\displaystyle {\begin{pmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&0&0\end{pmatrix}}}$